• #Lorentz transformations
  • #Rotation of coordinate system :-
  • #Pair of Events :-
• #Implications of Lorentz Transformation :-
  • #Simultanity is relative concept :
  • #Time dilation
• #Length contraction :-

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\begin{tikzpicture}
\draw[thick] (0,2.4) -- (0,0) -- (2.3,0);
\node[above right] at (0,2.4) {$S$};
\draw[thick] (2.3,2.4) -- (2.3,0) -- (8.4,0);
\node[above right] at (2.3,2.4) {$S'$};
\draw[->] (2.3,1.65) -- (2.95,1.65) node[right] {$u$};
\end{tikzpicture}

Event \((E_1)\): Fire cracker explode at \(x\) in frame \(S\) at time \(t\)

Event \((E_2)\): origins coincide at \(x=0\) at \(t=0\)

In \(S'\) frame \(E_1\) occurs at \((x',t')\)

\(E_2\) occurs at \((x'=0,\ t'=0)\)

\[ \text{Then} \quad x' = (x-ut)\,\gamma \qquad\qquad \text{where} \quad \gamma = \frac{1}{\sqrt{1-u^2/c^2}} \] \[ \text{and} \quad x = (x'+ut')\,\gamma \] \[ \text{and} \quad t' \;=\; \frac{x'}{c} \;=\; \frac{(x-ut)\,\gamma}{c} \;=\; \left( \frac{x}{c} - \frac{u}{c}\cdot t \right)\gamma \] \[ t' \;=\; \left( t - \frac{u}{c}\,\frac{x}{c} \right)\gamma \;=\; \left( t - \frac{ux}{c^{2}} \right)\gamma \] \[ \text{So,} \quad t' = \left( t - \frac{ux}{c^{2}} \right)\gamma \]

So coordinate of event \(E_1\) in frame \(S' = (x',t')\) where

\[ \boxed{\; \begin{aligned} x' &= (x-ut)\,\gamma\\ t' &= \left( t - \frac{ux}{c^{2}} \right)\gamma \end{aligned}\;} \qquad \text{where} \quad \gamma = \frac{1}{\sqrt{1-u^2/c^2}} \]

\begin{tikzpicture}[scale=1.15]
\draw[->] (-2.2,0) -- (2.7,0) node[right] {$x$};
\draw[->] (0,-1.8) -- (0,2.4) node[above] {$y$};
\draw[dashed,->] (-1.95,-1.37) -- (2.55,1.79) node[right] {$x'$};
\draw[dashed,->] (1.05,-1.50) -- (-1.45,2.07) node[above left] {$y'$};
\fill (1.20,1.70) circle (1.5pt);
\node[above right] at (1.15,1.72) {$(x,y)$};
\draw[dashed] (1.20,1.70) -- (1.20,0);
\draw[dashed] (1.20,1.70) -- (0,1.70);
\draw (0.55,0) arc (0:35:0.55);
\node at (0.78,0.20) {\scriptsize $\theta$};
\draw[<->] (0,-0.38) -- (1.20,-0.38);
\node[below] at (0.60,-0.36) {\scriptsize $x$};
\end{tikzpicture}

\[ \begin{aligned} x' &= x\cos\theta + y\sin\theta\\ y' &= -x\sin\theta + y\cos\theta \end{aligned} \]

Came from the way unit vectors transform

\begin{tikzpicture}
\draw[->] (0,0) -- (3.0,0);
\node[below] at (1.05,-0.02) {$\hat{e}_x$};
\draw[->] (0,0) -- (0,2.1);
\node[left] at (-0.05,1.25) {$\hat{e}_y$};
\draw[->] (0,0) -- (1.95,1.95);
\node[right] at (1.00,0.80) {$\hat{e}_x{}'$};
\draw[->] (0,0) -- (-1.40,1.40);
\node[left] at (-0.80,0.72) {$\hat{e}_y{}'$};
\draw (0.78,0) arc (0:45:0.78);
\node at (1.00,0.30) {\scriptsize $\theta$};
\draw (0,0.98) arc (90:135:0.98);
\node at (-0.22,1.10) {\scriptsize $\theta$};
\end{tikzpicture}

\[ \begin{aligned} \hat{e}_x{}' &= \hat{e}_x\cos\theta + \hat{e}_y\sin\theta\\ \hat{e}_y{}' &= -\hat{e}_x\sin\theta + \hat{e}_y\cos\theta \end{aligned} \]

The Lorentz transformation

\[ \left. \begin{aligned} x' &= \frac{x}{\sqrt{1-u^2/c^2}} \;-\; \frac{ut}{\sqrt{1-u^2/c^2}}\\ t' &= \frac{t}{\sqrt{1-u^2/c^2}} \;-\; \frac{ux/c^{2}}{\sqrt{1-u^2/c^2}} \end{aligned} \right\} \quad \begin{aligned} &\text{is } \underline{\text{not}}\\ &\text{just a rotation} \end{aligned} \]

\(E_1\): Bullet is fired at location \(x_1\) at time \(t_1\)

\(E_2\): bullet hits wall at location \(x_2\) at time \(t_2\)

If we observe same pair of events from a frame \(S'\) moving with speed `\(u\)' relative to \(S\).

\[ \begin{aligned} E_1 &: (x_1',\ t_1')\\ E_2 &: (x_2',\ t_2') \end{aligned} \]

Then from lorentz transformations :

\[ \begin{aligned} x_1' &= (x_1 - ut_1)\,\gamma\\ x_2' &= (x_2 - ut_2)\,\gamma \end{aligned} \qquad\qquad \begin{aligned} t_1' &= \left( t_1 - \frac{ux_1}{c^{2}} \right)\gamma\\ t_2' &= \left( t_2 - ux_2/c^{2} \right)\gamma \end{aligned} \]

lets find

\[ \begin{aligned} \Delta x' &= x_2' - x_1'\\ \Delta x' &= \left( x_2 - ut_2 - x_1 + ut_1 \right)\gamma\\ &= \left\{ (x_2-x_1) - u(t_2-t_1) \right\}\gamma\\ \Delta x' &= (\Delta x - u\,\Delta t)\,\gamma \end{aligned} \qquad \begin{aligned} \Delta t' &= t_2' - t_1'\\ &= \left( (t_2-t_1) - \frac{u(x_2-x_1)}{c^{2}} \right)\gamma\\ \Delta t' &= \left( \Delta t - \frac{u\,\Delta x}{c^{2}} \right)\gamma \end{aligned} \]

\(\Rightarrow\) Difference of coordinates & times also follows lorentz transformation

\[ \boxed{\; \begin{aligned} \Delta x' &= (\Delta x - u\,\Delta t)\,\gamma\\ \Delta t' &= \left( \Delta t - \frac{u\,\Delta x}{c^{2}} \right)\gamma \end{aligned}\;} \] \[ \text{velocity of bullet} \Big)_{S} \;=\; \frac{\Delta x}{\Delta t} \;=\; V \] \[ \text{velocity of bullet} \Big)_{S'} \;=\; \frac{\Delta x'}{\Delta t'} \;=\; W \]

from our results

\[ \frac{\Delta x'}{\Delta t'} \;=\; W \;=\; \frac{\Delta x - u\,\Delta t}{\Delta t - \dfrac{u\,\Delta x}{c^{2}}} \] \[ =\; \frac{\left( \dfrac{\Delta x}{\Delta t} - u \right)}{1 - \dfrac{u}{c^{2}}\left( \dfrac{\Delta x}{\Delta t} \right)} \] \[ \boxed{\; W \;=\; \frac{V-u}{1 - uV/c^{2}} \;} \]

So speed of bullet according to \(S' \neq V-u\) but

\[ W \;=\; \frac{V-u}{1-uV/c^{2}} \qquad (>\,V-u) \]

For Galilean result \(c\to\infty\)

\[ \Rightarrow \quad W \;=\; V\Big)_{S'} \;=\; \frac{V-u}{1-0} \;=\; V-u \] \[ \text{If} \quad V=c, \qquad W = \frac{c-u}{1 - \dfrac{u}{c}} \;=\; c \]

\begin{tikzpicture}
\draw[thick] (0,2.3) -- (0,0) -- (7.4,0);
\node[above right] at (0,2.3) {$S$};
\draw (1.7,2.3) -- (1.7,0.85) -- (6.9,0.85);
\node[above right] at (1.7,2.3) {$S'$};
\draw[->] (1.7,1.05) -- (2.3,1.05) node[right] {$u$};
\draw (3.9,1.12) circle (0.14);
\draw[->] (4.1,1.05) -- (4.75,1.05) node[right] {$W$};
\draw[->] (3.9,0.80) -- (3.9,0.98);
\end{tikzpicture}

What will be speed of bullet w.r.t. frame \(S\).

\[ \begin{aligned} \Delta x &= (\Delta x' + u\,\Delta t')\,\gamma\\ \Delta t &= \left( \Delta t' + \frac{u\,\Delta x'}{c^{2}} \right)\gamma \end{aligned} \] \[ \frac{\Delta x}{\Delta t} \;=\; V \;=\; \frac{\Delta x' + u\,\Delta t'}{\Delta t' + \dfrac{u\,\Delta x'}{c^{2}}} \] \[ V \;=\; \frac{\dfrac{\Delta x'}{\Delta t'} + u}{1 + \dfrac{u}{c^{2}}\,\dfrac{\Delta x'}{\Delta t'}} \] \[ \boxed{\; V \;=\; \frac{W+u}{1 + uW/c^{2}} \;} \]

So \(V \neq u+w\) but somewhat less than that

\[ \left\{\; V = \frac{W+u}{1+uW/c^{2}} \quad < \quad W+u \;\right\} \] \[ \text{If} \quad W=c, \qquad V = \frac{c+u}{1 + \dfrac{uc}{c^{2}}} \;=\; \left( \frac{c+u}{c+u} \right) c \;=\; c \] \[ \Delta t' = \left( \Delta t - \frac{u\,\Delta x}{c^{2}} \right)\gamma \]

If \(\Delta t = 0\) (simultaneous in `\(S\)' frame)

\[ \boxed{\; \Delta t' \;=\; -\,\frac{u\,\Delta x}{c^{2}}\,\gamma \;} \]

This implies that

\begin{tikzpicture}
\draw[thick] (0,1.95) -- (0,0.55) -- (10.4,0.55);
\node[above right] at (0,1.95) {$S$};
\draw (1.35,1.95) -- (1.35,1.18) -- (8.5,1.18);
\node[above right] at (1.35,1.95) {$S'$};
\draw[->] (1.35,1.52) -- (1.95,1.52) node[right] {$u$};
\draw (8.5,1.18) -- (8.5,2.00) -- (9.85,1.80) -- (9.85,1.10) -- cycle;
\node at (9.18,1.55) {\scriptsize engine};
\draw (2.85,1.05) circle (0.15);
\node[left] at (2.66,1.01) {\scriptsize clock};
\draw (6.95,1.05) circle (0.15);
\node[left] at (6.76,1.01) {\scriptsize clock};
\node at (3.85,1.62) {\scriptsize $(x_1',t_1')$};
\node at (7.75,1.62) {\scriptsize $(x_2',t_2')$};
\fill (3.25,0.55) circle (1.3pt);
\node[below,align=center] at (3.25,0.48) {$x_1$\\$t$};
\fill (7.25,0.55) circle (1.3pt);
\node[below,align=center] at (7.25,0.48) {$x_2$\\$t$};
\end{tikzpicture}

\[ \Rightarrow \quad \Delta t' \;=\; -\,\frac{u\,\Delta x}{c^{2}}\,\gamma \] \[ t_2' - t_1' \;=\; -\,\frac{u\,(x_2' - x_1')\,\gamma}{c^{2}} \] \[ \boxed{\; t_2' \;=\; t_1' \;-\; \frac{u\,\Delta x}{c^{2}}\,\gamma \;} \]

Clocks at frond end are behind the clocks of rear end by amount

\[ \frac{u\,\Delta x}{c^{2}}\,\gamma \;=\; \frac{u\,(\Delta x\,\gamma)}{c^{2}} \;=\; \frac{u\,L_0}{c^{2}} \] \[ (\Delta x)\,\gamma \;=\; \text{appeared seperation b/w events.} \] \[ \text{i.e.} \qquad \Delta x' = (\Delta x)\,\gamma \]

Or we can check,

\[ \Delta x' = (\Delta x - u\,\Delta t)\,\gamma \qquad (\Delta t = 0), \qquad \Delta x' = (\Delta x)\,\gamma \]

Or consider an example :-

\begin{tikzpicture}
\draw[thick] (0,2.4) -- (0,0) -- (10.2,0);
\draw (1.7,2.45) -- (1.7,1.45) -- (7.5,1.45) -- (7.5,1.05);
\node[above right] at (1.7,2.45) {$S'$};
\draw (1.50,1.28) -- (1.90,1.62);
\draw (1.50,1.62) -- (1.90,1.28);
\draw (1.70,1.22) -- (1.70,1.68);
\draw (7.30,1.28) -- (7.70,1.62);
\draw (7.30,1.62) -- (7.70,1.28);
\draw (7.50,1.22) -- (7.50,1.68);
\draw[->] (7.65,1.20) -- (8.35,1.20) node[right] {$u$};
\node at (3.55,1.15) {$c$};
\draw[<-] (3.80,1.15) -- (4.45,1.15);
\fill (4.60,1.15) circle (1.5pt);
\draw[->] (4.75,1.15) -- (5.40,1.15);
\node at (5.65,1.15) {$c$};
\end{tikzpicture}

w.r.t frame \(S'\) explosions are simultaneous

but w.r.t frame `\(S\)' explosions are not simultaneous.

\[ \Delta t = 0 \qquad\qquad \Delta t' \;=\; \frac{-\,u\,\Delta x}{c^{2}}\,\gamma \]

explosion at rear end happened earlier & explosion at frond end happens later.

\[ \begin{aligned} &\text{For}\\ &\text{Galilean}\\ &\underline{\text{Times}} \end{aligned} \left\{ \begin{aligned} &\text{If} \quad c\to\infty \qquad \Delta t' = \left( \Delta t - \frac{u\,\Delta x}{c^{2}} \right)\gamma\\ &\qquad\qquad\qquad \Delta t' = (\Delta t)\,\gamma \;=\; \frac{\Delta t}{\sqrt{1-u^2/c^2}} \;=\; \Delta t \end{aligned} \right. \]

(the \(u^2/c^2\) under the root goes to \(0\) as \(c\to\infty\))

(Q.1) I have a clock

\[ \begin{aligned} E_1 &: \text{Clock says tick at } x=0,\ t=0\\ E_2 &: \text{---------- tock at } x=0,\ t=\tau \end{aligned} \]

According to me

\[ \begin{aligned} \Delta t &= \tau - 0 = \tau\\ \Delta x &= 0 \end{aligned} \]

According to you who is in frame \(S'\) :-

\[ \Delta t' = \left( \Delta t - \frac{u\,\Delta x}{c^{2}} \right)\gamma \] \[ \boxed{\; \Delta t' = (\tau)\,\gamma \;} \] \[ \gamma > \underline{1} \qquad \Rightarrow \qquad (\Delta t' > \tau) \]

This is exactly what we expect, as according to \(S'\) `\(S\)' frame is moving so, \((\tau < \Delta t')\)

moving clocks runs slower.

If you are carrying clock and moving.

\[ \Delta t = \left( \Delta t' + \frac{u\,\Delta x'}{c^{2}} \right)\gamma \] \[ (\Delta t' = \tau, \qquad \Delta x' = 0) \] \[ \boxed{\; \Delta t \;=\; \frac{\Delta t'}{\sqrt{1-u^2/c^2}} \;=\; \frac{\tau}{\sqrt{1-u^2/c^2}} \;} \]

\((\Delta t > \Delta t')\) as expected. (first is \(I\), second is you)

I see your clock is moving, so \((\Delta t' < \Delta t)\).

It is similar to case, we both (I & you) carry a light clock.

\begin{tikzpicture}
\draw[thick] (0,3.5) -- (0,0) -- (9.2,0);
\draw (1.75,3.5) -- (1.75,1.80) -- (6.8,1.80);
\draw[->] (1.75,2.55) -- (2.40,2.55);
\node[right] at (2.42,2.55) {$u$};
\node at (2.45,2.05) {(You)};
\draw[->] (3.35,3.05) -- (4.70,1.88);
\draw[->] (4.70,1.88) -- (6.05,3.05);
\draw (3.05,3.02) -- (3.70,3.16);
\draw (5.75,3.02) -- (6.40,3.16);
\draw (4.35,1.80) -- (5.05,1.80);
\node at (0.95,0.80) {(I)};
\draw[->] (2.55,0.10) -- (2.55,1.30);
\draw (2.20,1.28) -- (2.90,1.42);
\end{tikzpicture}

We accuse each other of using slower clocks.

\begin{tikzpicture}
\draw[thick] (0,1.9) -- (0,0) -- (7.8,0);
\node[above left] at (0.1,1.9) {$S$};
\draw[->] (0,1.5) -- (0,1.9);
\draw (0.35,0.98) rectangle (3.45,1.36);
\node[above] at (1.90,1.38) {$L_0$};
\node at (1.90,1.16) {\scriptsize $S'$ frame};
\draw[->] (3.50,1.17) -- (4.25,1.17) node[right] {$u$};
\end{tikzpicture}

To measure length of rod in ground frame \((S)\) we need to mark ends of rod simultaneously \((\Delta t = 0)\).

\[ \begin{aligned} \Delta x &= \text{appeared length} = L\\ \Delta x' &= \text{Rest length} = L_0 \end{aligned} \] \[ \Delta x' \;=\; \frac{(\Delta x - u\,\Delta t)}{\sqrt{1-u^2/c^2}} \quad (\Delta t \to 0) \qquad \Rightarrow \qquad \Delta x \;=\; (\Delta x')\sqrt{1-u^2/c^2} \] \[ \boxed{\; L = L_0\sqrt{1-u^2/c^2} \;} \]

\(\left( \underline{\text{How}}\ ? \right)\) Actually the measurment of our rod in moving frame was not simultaneous

\[ \Delta t' = \left( \Delta t - \frac{u\,\Delta x}{c^{2}} \right)\gamma \quad (\Delta t \to 0) \] \[ \Delta t' \;=\; -\,\frac{u\,\Delta x}{c^{2}}\,\gamma \]

So You will claim that I measured front end first then rod passes some distance and then rear end marking were taken.

\begin{tikzpicture}
\draw (0.6,0.95) rectangle (3.4,1.38);
\node[above] at (2.0,1.40) {$1\,\mathrm{m}$};
\draw[->] (3.45,1.16) -- (4.20,1.16) node[right] {$u$};
\draw (5.3,0.95) -- (7.3,0.95);
\draw (8.3,0.95) -- (10.6,0.95);
\foreach \x in {5.4,5.65,...,7.2} { \draw (\x,0.95) -- (\x-0.18,0.72); }
\foreach \x in {8.4,8.65,...,10.5} { \draw (\x,0.95) -- (\x-0.18,0.72); }
\draw[<->] (7.3,0.55) -- (8.3,0.55);
\node[below] at (7.8,0.52) {$0.5\,\mathrm{m}$};
\node[below] at (7.8,0.12) {hole};
\end{tikzpicture}

(Q) A moving rod contracts to half of its length. will it fall ?

According to rod frame hole is just \(0.25\) m long so it might not fall.

But It will fall according to ground frame

what happens is --- Frond end enters in first, rear end comes later on. & rod falls in the hole.

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